I want show multiple selected languages from dropdown but it shows only one
if(isset($_POST['search'])){
$getuser = $user->search($_SESSION['id'], $_POST);
}
<select class="selectpicker btn btn-outline-light mx-0" data-width="fit" name="lang[]" multiple="" >
<?php while($row = $getlang->fetch_assoc()):; ?>
<option value="<?php echo $row['lang_id']; ?>">
<?php echo $row['name']; ?>
<?php endwhile ?>
</option>
</select>
<?php
public function search($id,$data){
$lang_id= $_REQUEST['lang'];
foreach($data['lang'] as $value) {
$q = "SELECT * FROM (
SELECT user_table.fname,user_table.lname, languages.name,languages.lang_id
FROM interested
INNER JOIN user_table ON interested.u_id = user_table.id
INNER JOIN languages ON interested.lang_id = languages.lang_id
WHERE NOT user_table.id='$id'
) AS I WHERE I.lang_id='$value' ";
}
$result = $this->connection->query($q);
return $result;
}
You can try to put the endwhile tag outside of the closing option tag. So your select option will look like this:
<select class="selectpicker btn btn-outline-light mx-0" data-width="fit" name="lang[]" multiple>
<?php while($row = $getlang->fetch_assoc()): ?>
<option value="<?php echo $row['lang_id']; ?>">
<?php echo $row['name']; ?>
</option>
<?php endwhile; ?>
</select>
Related
Why can't I get the value of the element? sorry I'm new at coding and trying to learn
<script>
function getdll() {
document.getElementById('lblmess').innerHTML =
(formid.listid[formid.listid.selectedIndex].value)
$element = document.getElementById("lblmess");
console.log($element.innerHTML)
}
</script>
<form name="formid">
<select class="custom-select" name="listid" onchange="getdll()">
<option value="0"></option>
<?php
$hall_qry = $conn->query("SELECT * FROM `office_hall` WHERE o_id !=0 order by `office_name` asc");
while ($row = $hall_qry->fetch_assoc()) : ?>
<option value="<?php echo $row['o_id'] ?>"><?php echo $row['office_name'] ?></option>
<?php
endwhile;
?>
</select>
<br><br>
<select class="custom-select">
<?php
$hall_qry = $conn->query("SELECT * FROM `assembly_hall` WHERE o_id = '$element' order by `room_name` asc");
while ($row = $hall_qry->fetch_assoc()) : ?>
<option value="<?php echo $row['id'] ?>"><?php echo $row['room_name'] ?></option>
<?php
endwhile;
?>
</select>
<label id="lblmess"></label>
</form>
this is where I use this code sorry if my coding id is like that .............................................................
Try:
onchange="getdll(this.value)"
THEN in your script, put a parameter
function getdll(val){
alert(val);
}
Should be able to get the value you selected
I want to echo the selected value from the database to update it then store it
for example I have an asset with category printers from table category which contains other categories and when I want to edit this asset on the edit page I should get a dropdown list contains all the categories and selected on printers then if I want to change it I will if not leave unchanged
The array is drop-down from table category inner joined with user_asset table in the database by asset_category as a foreign key
this is what I have done so far
<label for="basicinput">الصنف : </label>
<?php
$result = mysqli_query($conn, "SELECT * FROM category");
?>
<select name="asset_category" class="form-control" required>
<?php while( $row = mysqli_fetch_array($result)) {?>
<option value="<?php echo $row['category_id'];?>">
<?php echo $row['cate_name'];?>
</option>
<?php }?>
</select>
</div>
You can add if check if ($row['cate_name'] == 'computer') { ?> and then add selected to this option:
<label for="basicinput">الصنف : </label>
<?php
$result = mysqli_query($conn, "SELECT * FROM category");
?>
<select name="asset_category" class="form-control" required >
<?php while( $row = mysqli_fetch_array($result)) {
if ($row['cate_name'] == 'computer') { ?>
<option value="<?php echo $row['category_id'];?>" selected><?php echo $row['cate_name'];?></option>
<?php } else { ?>
<option value="<?php echo $row['category_id'];?>"><?php echo $row['cate_name'];?></option>
<?php }
}?>
</select>
Notice: If you have multiple elements with that category it will select the last one.
the answer is very simple.. let's put this code
<label for="basicinput">الصنف : </label>
<?php
$result = mysqli_query($conn, "SELECT * FROM category");
?>
<select name="asset_category" class="form-control" required>
<?php while( $row = mysqli_fetch_array($result)) {
if($row['cate_name']== printers) { ?>
<option value="<?php echo $row['category_id'];?>" selected="selected">
<?php echo $row['cate_name'];?> </option>
<?php } else { ?>
<option value="<?php echo $row['category_id'];?>">
<?php echo $row['cate_name'];?> </option>
<?php }?>
</select>
</div>
The logic is that using while loop, checking the condition using if class, and when it satisfies make it as selected. Then it will be echo as selected Value.
I need to get selected value from db into datalist box.Tell me how to do it. Here is the code.
<input list="Rank_Name" class="form-control" required>
<datalist id="Rank_Name">
<?php
$sel_cus = "select Rank_Name from ranks where Rank_Status=1";
$res_cus = mysqli_query($connection, $sel_cus);
while ($row = mysqli_fetch_array($res_cus)) {
?>
<option value="<?php echo $row['Rank_Name'];?>"></option>
<?php
}
?>
</datalist>
If i understood right , you need to select value in dropdownlist with other value also. You can achieve this by doing this
<?php
$select1="select Rank_Name from ranks where Rank_Status=1";
$q=mysqli_query($select1) or die($select1);
$row=mysqli_fetch_array($q); //here you are getting name of person whose rank is 1
?>
<datalist id="Rank_Name">
<?php
$s="select * from ranks ";
$q=mysqli_query($s) or die($s);
while($r=mysqli_fetch_array($q))
{ ?>
<option value="<?php echo $r['Rank_Name']; ?>"<?php if($row['Rank_Name']==$r['Rank_Name']) echo 'selected="selected"'; ?>>
<?php echo $r['Rank_Name']; ?>
</option>
<?php } ?>
</datalist>
In above code, this line <?php if($row['Rank_Name']==$r['Rank_Name']) echo 'selected="selected"'; ?> check if value are same from first query ,and if same then that option will be get selected automatically
<input list="Rank_Name" class="form-control" required>
<datalist id="Rank_Name">
<?php
$sel_cus = "select Rank_Name from ranks where Rank_Status=1";
$res_cus = mysqli_query($connection, $sel_cus);
while ($row = mysqli_fetch_array($res_cus)) {
echo "<option value=".$row['Rank_Name']."></option>";
}
?>
</datalist>
try this code.
im using echo <option> with while loop
I need to pull down a column of color codes for a color picker site I'm making but it won't work:
<select name="select" id="dropdown">
<option id="0">-- Select Color --</option>
<?php
$getColors = mysql_query("SELECT * FROM color_codes");
while($list = mysql_fetch_array($getColors)){
?>
<option id="<?php echo $list ['colorID']; ?>">
<?php echo $list['color_code'] ?></option>
<?php } ?>
</select>
Any ideas?
just did a couple of changes to your code and it worked for me, Hope it works for you.
<?php
$mysqli= new mysqli(DB_HOST,DB_USER,DB_PASS,DB_NAME) or die(mysqli_connect_error());
$sql="select * from test";
$result=$mysqli->query($sql);
?>
<select name="select" id="dropdown">
<option id="0">-- Select Color --</option>
<?php
while($list = $result->fetch_array(MYSQLI_ASSOC)){
?>
<option id="<?php echo $list['id']; ?>">
<?php echo $list['name'] ?></option>
<?php } ?>
</select>
I'm developing a webpage with a select list that contains images.
I already have this:
When I select an image name in the list the image will be displayed in the div below.
<?php
// Create connection
$con=mysqli_connect("******","***","***","charts");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
<form method="post" action="index.php" id="nano" name="nano">
<p>
<select name="SelectBox" id="SelectBox" onchange="this.form.submit()">
<?php if($_POST['submitted'] == true){ ?>
<?php
$result = mysqli_query($con,"SELECT * FROM Nano WHERE IMAGE_NAME ='". $_POST['SelectBox']."'");
while($row = mysqli_fetch_array($result))
{ ?>
<option selected="selected" value="<?php echo $row['IMAGE_NAME'] ?>">
<?php echo $row['IMAGE_PARAMETER'] ?>
</option>
<?php } ?>
<?php echo $_POST['SelectBox']; ?></option>
<?php } else{ ?>
<?php
$result = mysqli_query($con,"SELECT TOP * FROM Nano");
while($row = mysqli_fetch_array($result))
{
?>
<option selected="selected" value="<?php echo $row['IMAGE_NAME'] ?>">
<?php echo $row['IMAGE_PARAMETER'] ?>
</option>
<?php
$var1 = $row['IMAGE_NAME']; ?>
<?php
}
?>
<?php } ?>
<option value="" disabled="disabled"> -------- </option>
<?php
$result = mysqli_query($con,"SELECT * FROM Nano");
while($row = mysqli_fetch_array($result))
{ $values[] = $row['IMAGE_NAME'];
?>
<option value="<?php echo $row['IMAGE_NAME'] ?>">
<?php echo $row['IMAGE_PARAMETER'] ?>
</option>
<?php }?>
</select>
<input type="hidden" name="submitted" id="submitted" value="true" />
</p>
<?php if($_POST['submitted'] == true){ ?>
<p><img src="Images\Nano\<?php echo $_POST['SelectBox']?>" width="953" height="600" /></p>
<?php }else { ?>
<p><img src="Images\Nano\<?php print_r($values[0]) ?>" width="953" height="600" /></p>
<?php } mysqli_close($con);?>
</form>
</div>
I want when I move down in the select list the picture will change and not when I click on it in the select list.
In your case, you have to bind hover event to option, but there is no way to do what you want using native select control. The native one only answers when you click a different option from previous.
However, you can simulate a select control using html&css&js, that way when your cursor move down the simulated option(which might be a div or something), you can bind event handlers to it and display the name.